Civil Engineering Reference
In-Depth Information
y
S
y
G
t
z
S
z
x
F
IGURE
10.18
Shear of a
thin-walled closed section
beam
s
shown in Fig. 10.12. The equilibrium equation (Eq. (10.21)) is therefore applicable
and is
∂
q
∂
s
t
∂σ
x
∂
x
+
=
0
Substituting for
∂σ
x
/∂
x
from the derivation of Eq. (10.2) and integrating we obtain, in
an identical manner to that for an open section beam
s
s
S
y
I
zy
−
S
z
I
z
S
z
I
zy
−
S
y
I
y
q
s
=
tz
d
s
+
ty
d
s
+
q
s
,0
(10.24)
I
zy
I
zy
I
z
I
y
−
I
z
I
y
−
0
0
where
q
s
,0
is the value of shear flow at the origin of
s
.
It is clear from a comparison of Eqs (10.24) and (10.22) that the first two terms of the
right-hand side of Eq. (10.24) represent the shear flow distribution in an open section
beam with the shear loads applied through its shear centre. We shall denote this 'open
section' or 'basic' shear flow distribution by
q
b
and rewrite Eq. (10.24) as
q
s
=
q
b
+
q
s
,0
We obtain
q
b
by supposing that the closed section beam is 'cut' at some convenient
point, thereby producing an 'open section' beam as shown in Fig. 10.19(b); we take
the 'cut' as the origin for
s
. The shear flow distribution round this 'open section' beam
is given by Eq. (10.22), i.e.
s
s
S
y
I
zy
−
S
z
I
z
S
z
I
zy
−
S
y
I
y
q
b
=
tz
d
s
+
ty
d
s
I
zy
I
zy
I
z
I
y
−
I
z
I
y
−
0
0
Equation (10.22) is valid only if the shear loads produce no twist; in other words,
S
z
and
S
y
must be applied through the shear centre of the 'open section' beam. Thus by
'cutting' the closed section beam to determine
q
b
we are, in effect, transferring the line
of action of
S
z
and
S
y
to the shear centre,
S
s
,0
, of the resulting 'open section' beam.
The implication is, therefore, that when we 'cut' the section we must simultaneously
introduce a pure torque to compensate for the transference of
S
z
and
S
y
. We shall
show in Chapter 11 that the application of a pure torque to a closed section beam
results in a constant shear flow round the walls of the beam. In this case
q
s
,0
, which
